Attempt to prove that quotients of a solvable group are solvable

Question

Define $G$ to be a solvable group if there exists a subnormal series $1=N_0\unlhd N_1\unlhd N_2\unlhd \dots\unlhd N_{i-1}\unlhd N_i=G$ such that the factor groups $N_{j+1}/N_j$ are all abelian.

I want to prove that the quotient groups of $G$ are solvable.

Suppose $H\unlhd G$. We can construct a subgroup series
$1=HN_0/H\le HN_1/H\le HN_2/H\le \dots\le HN_{i-1}/H\le HN_i/H=G/H$. I know that $H\unlhd G$ implies $H\unlhd HN_j$, so the quotients are well-defined. Since $HN_j\le HN_{j+1}$, by the lattice isomorphism thm., $HN_j/H\le HN_{j+1}/H$

but I cannot figure out why $HN_j/H\unlhd HN_{j+1}/H$, i.e., why $HN_j\unlhd HN_{j+1}$. It becomes messy when I write something like $(h'n_{j+1})(hn_j)(h'n_{j+1})^{-1}$ and showing that it is indeed an element of $HN_j$. Why is it so?

Answer 1

Hint: proving this will simplify matters $$HN_j/H = \pi(N_j)$$ where $\pi: G \rightarrow G/H$ is the projection map.

I'll do the easy direction. $$\forall n \in N_j, nH = 1\cdot nH \in HN_j/H$$ So $\pi(N_j)\subseteq HN_j/H$

For the other direction, $$\forall hnH \in HN_j/H$$

Can you find a $n'\in N_j$ such that $n'H = hnH$?

Answer 2

From what you wrote it follows that we only need to show $\;HN_j\lhd HN_{j+1}\;$ , as then normality is preserved in the quotient group. But for any $\;h,h'\in H\;,\;\;x_k\in N_k$:

$$(hx_{j+1})^{-1}(h'x_j)(hx_{j+1})=x_{j+1}^{-1}(h^{-1}h'x_jh)x_{j+1}=x_{j+1}^{-1}(h^{-1}h'h)(h^{-1}x_jh)x_{j+1}\in$$

$$\in x_{j+1}^{-1} HN_jx_{j+1}=\left[x_{j+1}^{-1} Hx_{j+1}\right]\left[x_{j+1}^{-1}N_jx_{j+1}\right]\in HN_{j}$$

$$\begin{align*}\end{align*}$$